Question
Compute the smallest positive integer n for which
\sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} }
is an integer
Compute the smallest positive integer n for which
\sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} }
is an integer
Let
and
k = \sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} }
From the given condition, we know k is an integer and,
=\sqrt{10000-n}
Rearrange the above equation, we get
Square both (1) and (2), and addition of squared equations
Square (3) and subtract (5)
2ab = k^2-200
Then
Since ab>0, \dfrac{k^2}{2}-100>0, then k>10\sqrt{2}
Substituting (6) to (4) gives
Subtracting 2 times (6) from (5) gives
a^2+b^2-2ab = 200-k^2+200
(a-b)^2 = 400-k^2
Since a > b, we can cancel negative result when taking square root.
We can solve the equation system formed by (3) and (8) to determine a and b. But here we don't need to do so to answer the question.
In order for the equation (8) to be meaningful, the radicant must be greater than 0, that is
400-k^2>0
Solving the inequality gives
k<20
Since equation (7) is a decreasing function when k>10\sqrt{2}, the smallest n happens when k is the largest number. Exclude k=19 since it yields non integer for n. Therefore,
k = 18
Substituting (7) yields,
n = 10000-(\dfrac{18^2}{2} -100)^2 =10000-62^2 = 6156